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Category: Integration

find-f-a-b-0-cos-ax-cos-bx-x-2-a-2-x-2-b-2-dx-with-a-gt-0-and-b-gt-0-2-calculate-0-cos-x-cos-2x-x-2-1-x-2-4-dx-

Question Number 66466 by mathmax by abdo last updated on 15/Aug/19 $${find}\:\:{f}\left({a},{b}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{cos}\left({ax}\right){cos}\left({bx}\right)}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right)}{dx}\:\:{with}\:{a}>\mathrm{0}\:{and}\:{b}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left({x}\right){cos}\left(\mathrm{2}{x}\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)}{dx} \\…

1-calculate-by-residus-method-0-dx-1-x-2-3-2-find-the-value-of-0-1-1-x-4-1-x-2-3-dx-

Question Number 66459 by mathmax by abdo last updated on 15/Aug/19 $$\left.\mathrm{1}\right)\:{calculate}\:{by}\:{residus}\:{method}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}+{x}^{\mathrm{4}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }{dx} \\ $$…

nice-calculus-prove-that-1-0-1-li-2-1-x-2-pi-2-2-4-2-0-1-log-1-t-t-3-4-1-t-dt-pi-3-2-2-2-3-4-log-2-pi-2-hi

Question Number 131991 by mnjuly1970 last updated on 10/Feb/21 $$\:\:\:\:\:\:\:\:\:\:\:\:…{nice}\:\:\:\:\:\:\:\:\:{calculus}…\:\: \\ $$$$\:\:{prove}\:\:{that}\:: \\ $$$$\:\:\phi_{\mathrm{1}} =\int_{\mathrm{0}} ^{\:\mathrm{1}} {li}_{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\:−\mathrm{4} \\ $$$$\:\:\:\:\phi_{\mathrm{2}} =\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{log}\left(\mathrm{1}−{t}\right)}{{t}^{\frac{\mathrm{3}}{\mathrm{4}}}…

Determine-the-results-of-the-following-I-1-a-e-x-a-e-x-dx-I-2-tanx-tanx-1-tan-2-x-dx-

Question Number 907 by Yugi last updated on 19/Apr/15 $${Determine}\:{the}\:{results}\:{of}\:{the}\:{following}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{I}_{\mathrm{1}} =\int\sqrt{\frac{{a}+{e}^{{x}} }{{a}−{e}^{{x}} }}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{I}_{\mathrm{2}} =\int\frac{\left({tanx}\right)\mid{tanx}\mid}{\mathrm{1}−{tan}^{\mathrm{2}} {x}}{dx} \\ $$ Commented by prakash jain…